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JEE Main 20266 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2026 · 6 April · Evening Shift · Chemistry · Solutions

Given below are two statements : Statement I : H _2 O molecules...

Given below are two statements :
Given below are two statements : Statement I : H _2 O molecules move from the chamber 1 to chamber 2 . Statement II :
Statement I : H2_2O molecules move from the chamber 11 to chamber 22.
Statement II : The osmotic pressure of a solution prepared by dissolving 5050 mg of potassium sulphate (molar mass =174= 174 g/mol) in 22 L of water (at 27°27°C) is 0.01070.0107 bar. (Given : R =0.083= 0.083 dm3^3 bar K−1^{-1} mol−1^{-1} and assume complete dissociation of electrolyte)
In the light of the above statements, choose the correct answer from the options given below :

Options

  1. ABoth Statement I and Statement II are true
  2. BBoth Statement I and Statement II are false
  3. CStatement I is true but Statement II is false
  4. DStatement I is false but Statement II is true

Correct answer

D. Statement I is false but Statement II is true

Step-by-step solution

For Statement I:
Molarity of the solution in Chamber 1 (M1M_1) = 18/180100/1000=1\dfrac{18 / 180}{100 / 1000} = 1 M
Molarity of the solution in Chamber 2 (M2M_2) = 30/180250/1000=1/61/4=0.667\dfrac{30 / 180}{250 / 1000} = \dfrac{1/6}{1/4} = 0.667 M
Since M1>M2M_1 > M_2, the solution in Chamber 1 has a higher solute concentration. During osmosis, the solvent (water) moves from the region of lower solute concentration to the region of higher solute concentration across a semi-permeable membrane. Thus, water will move from Chamber 2 to Chamber 1. Statement I is false.

For Statement II:
The osmotic pressure is given by π=iCRT\pi = iCRT
For K2SO4K_2SO_4, assuming complete dissociation (K2SO4→2K++SO42−K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}), the van't Hoff factor i=3i = 3.
Concentration C=50×10−3174×2C = \dfrac{50 \times 10^{-3}}{174 \times 2} mol/L
π=3×(50×10−3174×2)×0.083×300\pi = 3 \times \left( \dfrac{50 \times 10^{-3}}{174 \times 2} \right) \times 0.083 \times 300
π=3×0.05×24.9348=3.735348≈0.0107\pi = \dfrac{3 \times 0.05 \times 24.9}{348} = \dfrac{3.735}{348} \approx 0.0107 bar
Statement II is true.

Therefore, Statement I is false but Statement II is true.

Answer: Statement I is false but Statement II is true
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